Official Quant thread for CAT 2013

@amresh_maverick said:
@scrabblerIn how many ways can four persons be selected from ten persons sitting around a circular table such that none of the selected person were sitting next to each other previously?A. 10 B. 15 C. 25 D. 100
I make it 25. I actually enumerated :P

Let us call the ten people 1, 2, 3, 4, 5...10

Consider when 1 is selected (here 10 can't be selected):
1, 3, 5, ke saath 789 (3 cases)
1, 3, 6 ke saath 89 (2 cases)
1, 3, 7 (1 case)
1, 4, 6 (2)
1, 4, 7 (1)
1, 5, 7 (1)

Now if 1 not chosen, then starting with 2 (here 10 can be selected)
again we find
2, 4, 6 (3 cases)
2, 4, 7 (2)
2, 4, 8 (1)
2, 5, 7 (2)
2, 5, 8 (1)
2, 6, 8(1)

With 3
3, 5, 7 (2)
3, 5, 8 (1)
3, 6, 8 (1)

With 4

4, 6, 8 (1)

Total 25. :splat: I am sure there must be an easier way but not seeing it right now...

regards
scrabbler
10*5C3/ 4 = 25 - kya hai ye :)


@scrabbler
@amresh_maverick said:
10*5C3/ 4 = 25 - kya hai ye @scrabbler
No clue 😃 at 2 am, I am not even trying to figure it out ;)

regards
scrabbler
@scrabbler only one number:- 2
@Akrosh007 said:
it's 3/8.
Can you please explain?
@amresh_maverick said:
10*5C3/ 4 = 25 - kya hai ye @scrabbler
what's logic behind this expression?
@amresh_maverick said:
@scrabblerIn how many ways can four persons be selected from ten persons sitting around a circular table such that none of the selected person were sitting next to each other previously?A. 10 B. 15 C. 25 D. 100
Select 1 person among 10 -> 10 ways

Select 3 person among remaining 9 such that none are adjacent -> this is now same as selecting 3 person out of 9 in linear arrangement, such that none are adjacent

So this can be done in C(5,3) ways (same as placing 3 people between the 6 where 6 people have 5 gaps between them)

So total ways = 10*C(5,3)

Now each quadruple has been repeated 4 times since we have 4 people that we have picked up.

So total unique cases = 10*C(5,3)/4


One can generalize this solution in that if we are asked to form a polygon of n 'sides' using the vertices of a larger polygon of N sides such that no two vertices are adjacent, then total ways = N/n * C(N-n-1, n-1)

@jaspunit said:
(sin @ +sin 2@+sin3@-sin4@)/(cos@+cos2@+cos3@+cos4@)Answer in terms of tan
Put @ = 30 deg

then, the Exp = _/3

So, tan(@) = 1/_/3

and tan(2@) = _/3

So, Exp = tan(2@)

Two players play a game using the interval [0,33] on the x-axis. The first player randomly chooses a square of side length s∈Z+, which has a side that lies entirely on the interval. The second player randomly chooses a circle with radius r∈Z+, which has a diameter that lies entirely on the interval. After repeating choosing random squares and circles in this fashion, the players realize that the probability that the circle and square intersect is 1/2. Let S={(s,r):p probability of intersection is 12}. Determine ∑(s,r)∈S {(s+r).}


Clarification of notation: The set S is the set of all ordered pairs of integers, (s,r), such that the probability that a square of side length sand a circle of radius r will intersect is 1/2.
What is minimum value when three values are added??
Does the minimum value occur at a point where all entities added are equal?
What is rule of symmetry in maxima and minima?
Please shed some light on these above topics pleaseeeee??

@Tusharrr
@jaspunit
@Aizen
@Angadbir
@mani0303
@neelshah3192

a girl has to climb 12 steps .she climbs in either a single step or 2 steps simultaneously .in how many ways can she do it ?

@amresh_maverick said:
@scrabblerIn how many ways can four persons be selected from ten persons sitting around a circular table such that none of the selected person were sitting next to each other previously?A. 10 B. 15 C. 25 D. 100

open the circle making a straight line

x1 - x2 - x3 - x4 -x5

x1 and x5 can be 0 but x2 x3 x4 can not

x1+x2+x3+x4+x5 = 6 => 7c4 = 35 ways


but we need to exclude cases when both persons at extreme ends selected, because when it makes circle these persons will be adjasent
in this case 2 people adjascent to erxtreme end can not be selected
so we left with 6 option
out of these 2 people are to be selected who are not adjasent

x1-x2-x3 (x1 x3 can be 0 but x2 can not)

x1+x2+x3 = 4 => 5c2 = 10

35 - 10 = 25 ways

question 2 ..
The number of ways in which we can select 5 numbers from the set of numbers(1,2,3,4,5.......25) such that none of the selection include any two consecutive number?

@fireatwill : its x + 2y = 12. Substitute values for x and y.
Answer: 7 ways
At the end of MAHABHARAT, Pandav's family is saying goodbye to Kaurav's family. Each member of Pandav's family says goodbye to each member of Kaurav's family. To say goodbye, two men shake hands and a man and a woman and two women kiss once on the cheek. An eyewitness to the event counted 21 handshakes and 34 kisses. If there are at least one man and one woman in each family, then how many MEN and WOMEN were there?
@vampgm BRO ANSWER IS 233

QUESTION 3
the total number of seven digit numbers, the sum of whose digits is even?

@erm total men are 5 pandavas, Vidur, dhritrashtra,Sanjay
Women are Draupadi, Kunti, Gandhari, Subhadra, Vidur's Wife,Uttara

will add more once i remember..
What is the remainder if the sum 2012^2013+2013^2012 is divided by 2012.2013?
@erm said:
At the end of MAHABHARAT, Pandav's family is saying goodbye to Kaurav's family. Each member of Pandav's family says goodbye to each member of Kaurav's family. To say goodbye, two men shake hands and a man and a woman and two women kiss once on the cheek. An eyewitness to the event counted 21 handshakes and 34 kisses. If there are at least one man and one woman in each family, then how many MEN and WOMEN were there?
suppose a men in Pandavas family and b men in Kaurav's family
c women in Pandav's family and d women in Kaurav's family


a*b = 21 => 3*7 or 1*21

a*d + b*c + c*d = 34

a*d + c*(b+d) = 34

a = 3, b = 7, c = 2, d = 4 satisfies
a = 7, b = 3, c = 4, d = 2 satisfies

in both cases, 10 Men and 6 women